I want to know about the latest cutoff information for B.Sc. in Computer Science at Mar Gregorios College.
Asked about Mar Gregorios College - B.Sc. in Computer Science
-
1 Answer
-
A score of 50% to 60% in Class 12 higher secondary coursework from an accredited board is required for a BSc in Computer Science
Students must have obtained a minimum of 45% marks in aggregate (40% for SC/ST/OBC) in PCM. The candidate must have also written any one of the following entrance exams: CET/ COMED-K / JEE / AIEEE. The rank card has to be produced.
The required percentage for similar programs like BSc Computer Science or Bachelor of Computer Applications (BCA) can vary based on the institution. Generally, a minimum of around 50% to 60% in Class 12 is often expected for Computer Science admission in 2024
...more
Similar Questions for you
Presented below is a comparison between the BCom LLB courses at Kerala University and Mar Gregorios College, based on seats, and Shiksha rating:
College/University | Shiksha Rating | Seats |
---|---|---|
BCom LLB at Kerala University | 3.5/5 | 132 |
BCom LLB at Gregorios College | 4.6/5 | 66 |
Note: The above information is taken from various official sources. Hence, it is indicative.
To know which college to choose, students compare both colleges based on important parameters such as fees, rating, and number of students. To help make the choice, presented below is a comparison between the BA LLB courses at Kerala University and Mar Gregorios College, based on the number of seats, fees, and Shiksha rating:
College/University | Fees | Shiksha Rating | Seats |
---|---|---|---|
BA LLB at Kerala University | INR 2.1 lakh | 3.5/5 | 132 |
BA LLB at Gregorios College | INR 2.21 Lakh | 4.6/5 | 66 |
Note: The above information is taken from various official sources. Hence, it is indicative.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers